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Ehrenfest's theorem

A simple way to calculate the expectation value of momentum is to evaluate the time derivative of $\langle x\rangle$, and then multiply by the mass $m$: i.e.,
\begin{displaymath}
\langle p \rangle = m\,\frac{d\langle x\rangle}{dt} = m\,\f...
...\infty}x\,\frac{\partial\vert\psi\vert^{\,2}}{\partial t}\,dx.
\end{displaymath} (152)

However, it is easily demonstrated that
\begin{displaymath}
\frac{\partial\vert\psi\vert^{\,2}}{\partial t} + \frac{\partial j}{\partial x} = 0
\end{displaymath} (153)

[this is just the differential form of Eq. (136)], where $j$ is the probability current defined in Eq. (137). Thus,
\begin{displaymath}
\langle p\rangle = -m\int_{-\infty}^{\infty} x\,\frac{\partial j}{\partial x}\,dx
= m\int_{-\infty}^{\infty}j\,dx,
\end{displaymath} (154)

where we have integrated by parts. It follows from Eq. (137) that
\begin{displaymath}
\langle p\rangle = - \frac{{\rm i}\,\hbar}{2}\int_{-\infty}^...
...nfty}^{\infty}
\psi^\ast\,\frac{\partial\psi}{\partial x}\,dx,
\end{displaymath} (155)

where we have again integrated by parts. Hence, the expectation value of the momentum can be written
\begin{displaymath}
\langle p\rangle = m\,\frac{d\langle x\rangle}{dt}= - {\rm i...
...nfty}^{\infty}
\psi^\ast\,\frac{\partial\psi}{\partial x}\,dx.
\end{displaymath} (156)

It follows from the above that

$\displaystyle \frac{d\langle p\rangle}{dt}$ $\textstyle =$ $\displaystyle -{\rm i}\,\hbar\int_{-\infty}^{\infty}
\left(\frac{\partial\psi^\...
...}{\partial x} + \psi^\ast\,\frac{\partial^2\psi}{\partial t\partial x}\right)dx$  
  $\textstyle =$ $\displaystyle \int_{-\infty}^{\infty}\left[ \left({\rm i}\,\hbar\,\frac{\partia...
...artial x}\left({\rm i}\,\hbar\,\frac{\partial\psi}{\partial t}\right)\right]dx,$ (157)

where we have integrated by parts. Substituting from Schrödinger's equation (119), and simplifying, we obtain
\begin{displaymath}
\frac{d\langle p\rangle}{dt} = \int_{-\infty}^{\infty}
\left...
...y} V(x)\,\frac{\partial \vert\psi\vert^{\,2}}{\partial x}\,dx.
\end{displaymath} (158)

Integration by parts yields
\begin{displaymath}
\frac{d\langle p\rangle}{dt} =-\int_{-\infty}^{\infty} \frac...
...psi\vert^{\,2}\,dx =- \left\langle \frac{dV}{dx}\right\rangle.
\end{displaymath} (159)

Hence, according to Eqs. (156) and (159),

$\displaystyle m\,\frac{d\langle x\rangle}{dt}$ $\textstyle =$ $\displaystyle \langle p\rangle,$ (160)
$\displaystyle \frac{d\langle p\rangle}{dt}$ $\textstyle =$ $\displaystyle -\left\langle \frac{dV}{dx}\right\rangle.$ (161)

Clearly, the expectation values of displacement and momentum obey time evolution equations which are analogous to those of classical mechanics. This result is known as Ehrenfest's theorem.

Suppose that the potential $V(x)$ is slowly varying. In this case, we can expand $dV/dx$ as a Taylor series about $\langle x\rangle$. Keeping terms up to second order, we obtain

\begin{displaymath}
\frac{dV(x)}{dx} = \frac{dV(\langle x\rangle)}{d\langle x\ra...
...e x\rangle)}{d\langle x\rangle^3}\,(x-\langle x\rangle)^{\,2}.
\end{displaymath} (162)

Substitution of the above expansion into Eq. (161) yields
\begin{displaymath}
\frac{d\langle p\rangle}{dt} = - \frac{dV(\langle x\rangle)}...
...{\,2}}{2}\,\frac{dV^3(\langle x\rangle)}{d\langle x\rangle^3},
\end{displaymath} (163)

since $\langle 1\rangle =1$, and $\langle x-\langle x\rangle\rangle = 0$, and $\langle (x-\langle x\rangle)^{\,2}\rangle = \sigma_x^{\,2}$. The final term on the right-hand side of the above equation can be neglected when the spatial extent of the particle wave-function, $\sigma_x$, is much smaller than the variation length-scale of the potential. In this case, Eqs. (160) and (161) reduce to
$\displaystyle m\,\frac{d\langle x\rangle}{dt}$ $\textstyle =$ $\displaystyle \langle p\rangle,$ (164)
$\displaystyle \frac{d\langle p\rangle}{dt}$ $\textstyle =$ $\displaystyle -\frac{dV(\langle x\rangle)}{d\langle x\rangle}.$ (165)

These equations are exactly equivalent to the equations of classical mechanics, with $\langle x\rangle$ playing the role of the particle displacement. Of course, if the spatial extent of the wave-function is negligible then a measurement of $x$ is almost certain to yield a result which lies very close to $\langle x\rangle$. Hence, we conclude that quantum mechanics corresponds to classical mechanics in the limit that the spatial extent of the wave-function (which is typically of order the de Boglie wave-length) is negligible. This is an important result, since we know that classical mechanics gives the correct answer in this limit.


next up previous contents
Next: Operators Up: Fundamentals of quantum mechanics Previous: Expectation values and variances   Contents
Richard Fitzpatrick 2006-12-12